= Solution
Expand about $t=t_{n+2}$ and use $y''=f'(y)f(y)$ from the <chain rule>. The residual of the exact solution is
$$
\begin{aligned}
&y(t)-\frac87y(t-h)+\frac17y(t-2h)-\frac67hy'(t)+\frac27h^2y''(t)\\
&\qquad=\frac1{21}h^4y^{(4)}(t)+O(h^5).
\end{aligned}
$$
The coefficients through $h^3$ vanish, and the displayed fourth-order coefficient does not. Thus the <multiderivative multistep method> has \b[order three].
At $h=0$, its first characteristic polynomial is
$$
\rho(\xi)=\xi^2-\frac87\xi+\frac17=(\xi-1)(\xi-1/7).
$$
Its roots are $1$ and $1/7$, with the unit-modulus root simple, so it satisfies the <root condition for a multistep method> and is <zero-stable>. Combining this with the defect estimate proves \b[convergence of order three], assuming sufficiently smooth $f$, order-three starting values, and the nearby branch of each implicit update.
More explicitly, the right-hand side is $hF_h(y_{n+2})$, where $F_h=6f/7-2hf'f/7$ has a uniform local <Lipschitz continuity> bound for small $h$. A bounded zero-stable impulse response and the <discrete Gronwall inequality> therefore control the accumulated $O(h^4)$ defects by $O(h^3)$ over fixed time intervals. This is <convergence of a zero-stable multiderivative method>, rather than a direct application of the first-derivative-only <Dahlquist equivalence theorem>.
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